WEBVTT
00:00:02.110 --> 00:00:06.760
In this video, we’re going to learn about the phenomenon of resonance in alternating-current circuits.
00:00:07.090 --> 00:00:13.760
Resonance occurs because both the capacitive and inductive reactances depend on the frequency of the alternating voltage and current.
00:00:13.990 --> 00:00:22.370
Let’s start by reviewing reactance, which generalizes the current-opposing quality of resistance for resistors to include inductors and capacitors as well.
00:00:22.750 --> 00:00:26.950
A resistor has the special property that its opposition to current is fixed.
00:00:27.170 --> 00:00:34.960
That is, the resistor’s resistance, usually given the symbol 𝑅, is unaffected by the strength, direction, or frequency of the voltage in the circuit.
00:00:35.250 --> 00:00:39.560
The same is not true of inductors and capacitors in alternating-current circuits.
00:00:39.760 --> 00:00:49.090
Although the capacitance 𝐶 and the inductance 𝐿 don’t depend on the voltage, the inductor and capacitor’s opposition to current does depend on the frequency of the voltage in the circuit.
00:00:49.330 --> 00:00:53.240
For a capacitor, the more charged it is, the more it opposes current.
00:00:53.450 --> 00:01:00.170
The faster the electromotive force changes directions, that is, the higher its frequency, the less the capacitor charges before discharging again.
00:01:00.610 --> 00:01:04.320
So at higher frequencies, the capacitor has a smaller reactance.
00:01:04.600 --> 00:01:07.310
An inductor, on the other hand, builds up a magnetic field.
00:01:07.460 --> 00:01:11.290
And the stronger the magnetic field gets, the less the inductor opposes current.
00:01:11.580 --> 00:01:14.060
However, this magnetic field takes time to build up.
00:01:14.270 --> 00:01:19.870
So the higher the frequency of the electromotive force, the weaker the magnetic field will be before changing directions.
00:01:20.330 --> 00:01:24.120
As a result, the inductive reactance will be larger at higher frequencies.
00:01:24.370 --> 00:01:30.810
As formulas, the capacitive reactance is one divided by the angular frequency of the voltage and current times the capacitance.
00:01:31.030 --> 00:01:35.870
The inductive reactance is the angular frequency of the voltage and current times the inductance.
00:01:36.090 --> 00:01:40.950
Note that both of these formulas give the correct qualitative relationship between reactance and frequency.
00:01:41.180 --> 00:01:47.100
The capacitive reactance is inversely proportional to frequency, while the inductive reactance is directly proportional to frequency.
00:01:47.320 --> 00:01:52.250
So at higher frequencies, the inductive reactance is larger and the capacitive reactance is smaller.
00:01:52.800 --> 00:01:58.480
𝜔, the angular frequency, is defined as two 𝜋 radians times the regular frequency, or cycles per second.
00:01:58.930 --> 00:02:04.930
We use 𝜔 because it helps us to write out these formulas simply without having to carry around factors of two 𝜋.
00:02:05.010 --> 00:02:13.700
The last thing we need to recall is that for a circuit with both inductive and capacitive components, the total reactance is not simply the sum of the inductive and capacitive reactances.
00:02:14.240 --> 00:02:19.430
This is because inductors and capacitors also introduce a phase shift between the current and the emf.
00:02:19.750 --> 00:02:25.000
Capacitors cause the current to leave the emf, while inductors cause the current to lag the emf.
00:02:25.230 --> 00:02:35.630
The net effect of these different phase shifts is that the correct combination for total reactance is the difference between the two reactances, inductive reactance minus capacitive reactance.
00:02:36.040 --> 00:02:42.240
Resonance will be possible in an alternating-current circuit precisely because total reactance is a difference instead of a sum.
00:02:42.840 --> 00:02:47.240
So let’s see how the difference between the inductive and capacitive reactances can lead to resonance.
00:02:47.490 --> 00:02:54.020
Let’s consider a simple circuit driven by an alternating-voltage source with an inductor and capacitor connected in series.
00:02:54.270 --> 00:02:59.580
Although we’ll limit our discussion to series circuits, the same principles still apply to parallel circuits.
00:02:59.800 --> 00:03:05.040
At any rate, the total reactance in the circuit is the inductive reactance minus the capacitive reactance.
00:03:05.180 --> 00:03:09.540
Remember that the inductive and capacitive reactances depend on frequency in opposite ways.
00:03:09.680 --> 00:03:16.760
So if we vary the frequency from, say, very low value to very high values, the capacitive reactance will change from very large to very small.
00:03:16.920 --> 00:03:20.080
But the inductive reactance will change from very small to very large.
00:03:20.390 --> 00:03:26.780
This suggests that there may be some frequency in the middle of our range for which the inductive reactance and the capacitive reactance are equal.
00:03:27.200 --> 00:03:32.380
If the inductive and capacitive reactances are equal, then their difference, the total reactance, is zero.
00:03:32.590 --> 00:03:37.470
So the inductor and capacitor combination provides no opposition to current at that frequency.
00:03:37.840 --> 00:03:44.230
To find the special frequency, we’ll start by equating the frequency-dependent formulas for inductive and capacitive reactance.
00:03:44.530 --> 00:03:48.260
To solve for 𝜔, we’ll multiply both sides by 𝜔 over 𝐿.
00:03:48.550 --> 00:03:52.430
On the left-hand side, the 𝐿 from our formula cancels the 𝐿 in the denominator.
00:03:52.490 --> 00:03:56.600
And on the right-hand side, the 𝜔 from our formula cancels the 𝜔 in the numerator.
00:03:57.090 --> 00:04:00.940
This leaves us with 𝜔 squared is equal to one divided by 𝐿𝐶.
00:04:01.160 --> 00:04:13.630
If we now go ahead and take the square root of both sides of this equality, we get that the angular frequency versus the total reactance is zero is equal to one divided by the square root of the inductance of the inductor times the capacitance of the capacitor.
00:04:14.020 --> 00:04:18.250
We often write 𝜔 with a subscript of zero when referring specifically to this frequency.
00:04:18.650 --> 00:04:24.640
The phenomenon of inductive and capacitive reactances exactly canceling at a particular frequency is known as resonance.
00:04:24.830 --> 00:04:28.920
And 𝜔 naught, the frequency at which resonance occurs, is called the resonant frequency.
00:04:29.440 --> 00:04:35.660
Because on resonance the total reactance is zero, the inductor and capacitor in our circuit behave like conducting wires.
00:04:35.830 --> 00:04:40.270
But this means that on resonance, the ideal circuit that we’ve drawn is effectively a short circuit.
00:04:40.630 --> 00:04:46.450
If this were a real circuit, a short circuit could cause serious damage to the various components and the AC voltage source.
00:04:46.600 --> 00:04:52.080
However, if this were a real circuit, there’ll be some resistance inherent in the components and in the wires.
00:04:52.320 --> 00:04:58.700
So let’s model this real-life situation with an ideal circuit that consists of an inductor, a capacitor, and also a resistor.
00:04:59.230 --> 00:05:05.590
Here, we have a circuit with a resistor, inductor, and capacitor, all in series, driven by an alternating-voltage source.
00:05:05.800 --> 00:05:14.810
Since this circuit has both resistive and reactive elements, the total opposition to current is given by the combination of resistance and reactance, known as impedance.
00:05:15.160 --> 00:05:20.760
The size of the impedance is the square root of the sum of the squares of the resistance and the total reactance.
00:05:20.910 --> 00:05:27.980
We have to use this special combination because reactive components change the phase between emf and current, but resistive components don’t.
00:05:28.300 --> 00:05:32.750
Okay, so let’s see what happens to the impedance when we drive the circuit at the resonant frequency.
00:05:33.110 --> 00:05:36.810
Recall that on resonance, the inductive and capacitive reactances are equal.
00:05:36.950 --> 00:05:38.830
So the total reactance is zero.
00:05:39.130 --> 00:05:45.790
So the impedance on resonance is the square root of 𝑅 squared plus zero, which is the square root of 𝑅 squared, which is just 𝑅.
00:05:46.100 --> 00:05:50.290
So on resonance, the impedance of the circuit is identically the resistance.
00:05:50.510 --> 00:05:52.340
This tells us several important things.
00:05:52.420 --> 00:06:00.210
First, since the total reactive contribution to impedance is zero on resonance, there is no phase shift introduced between the current and the emf.
00:06:00.460 --> 00:06:06.480
Secondly, for a general alternating-current circuit, Ohm’s law tells us that voltage is equal to current times impedance.
00:06:06.590 --> 00:06:14.120
On resonance, this becomes voltage is equal to current times resistance, which is just Ohm’s law for purely resistive alternating-current circuits.
00:06:14.300 --> 00:06:20.580
Furthermore, looking back at our formula for the size of impedance, total reactance squared is always positive unless it’s zero.
00:06:20.930 --> 00:06:25.680
So on resonance, when total reactance is zero, the impedance is minimized.
00:06:25.890 --> 00:06:32.130
Looking back at Ohm’s law, if the peak voltage doesn’t change, then as impedance gets smaller, peak current gets larger.
00:06:32.520 --> 00:06:35.680
So a minimum impedance implies a maximum peak current.
00:06:35.830 --> 00:06:43.210
Now that we’ve seen what happens when we drive the circuit at the resonant frequency, let’s see what happens when we drive the circuit at frequencies other than the resonant frequency.
00:06:43.430 --> 00:06:53.640
To see how the current behaves at frequencies other than the resonant frequency, we use a graph with angular frequency on the horizontal axis and relative amplitude of the current on the vertical axis.
00:06:53.860 --> 00:07:02.530
The relative amplitude for the current at a particular angular frequency is found by dividing the peak current at that frequency by the peak current at the resonant frequency.
00:07:02.870 --> 00:07:07.260
So, by definition, the relative amplitude for the current at the resonant frequency is one.
00:07:07.660 --> 00:07:16.370
If at some other frequency the current had a peak value that was half the value of the current at the resonant frequency, then the relative amplitude of the current at that frequency would be one-half.
00:07:16.760 --> 00:07:23.490
Using relative instead of absolute amplitude allows this discussion to be very general so it applies to a wide variety of resonant phenomena.
00:07:23.860 --> 00:07:31.760
Turning back in particular to our electronic circuits at frequencies increasingly large relative to the resonant frequency, the inductive reactance is larger and larger.
00:07:31.980 --> 00:07:34.630
And so the relative amplitude of the current is smaller and smaller.
00:07:35.150 --> 00:07:41.300
Similarly, at frequencies increasingly small relative to the resonant frequency, the capacitive reactance is larger and larger.
00:07:41.390 --> 00:07:44.720
And so, again, the relative amplitude of the current is smaller and smaller.
00:07:45.190 --> 00:07:49.920
This graph that we’ve drawn actually has a shape that’s typical for a wide variety of resonant systems.
00:07:50.060 --> 00:07:54.140
One of the most striking features of this graph is the sharp peak at the resonant frequency.
00:07:54.430 --> 00:07:58.260
We say that the peak is sharp because it is much narrower than it is tall.
00:07:58.730 --> 00:08:07.740
What this means physically is that the current in our circuit will be much larger when driven at the resonant frequency than when driven at frequencies not more smaller or larger than the resonant frequency.
00:08:08.350 --> 00:08:14.830
For many applications, from measurement equipment to radio communications, it’s useful to quantify the sharpness of the resonant peak.
00:08:15.150 --> 00:08:21.150
This is because the sharper the resident peak, the more selectively our system responds strongly at a particular frequency.
00:08:21.480 --> 00:08:27.670
Equivalently, the sharper the peak, the more our systems respond to changes for smaller shifts away from the resonant frequency.
00:08:27.860 --> 00:08:33.750
The number that we use to quantify the sharpness of the peak is called the 𝑄- or quality factor of the resonance.
00:08:33.830 --> 00:08:42.940
For series circuits like the kind we’ve been considering, the 𝑄-factor is equal to the angular frequency of the resonance times the inductance of the inductor divided by the resistance of the resistor.
00:08:43.480 --> 00:08:46.000
There are a number of other ways we could define the 𝑄-factor.
00:08:46.170 --> 00:08:51.960
No matter how we define it though, larger 𝑄-factors correspond to graphs that are more sharply peaked around the resonant frequency.
00:08:52.120 --> 00:08:56.670
And smaller 𝑄-factors correspond to graphs that are more broadly spread out around the resonant frequency.
00:08:57.280 --> 00:09:04.350
In fact, it turns out that the width of the peak at about half the maximum value is approximately the resonant frequency divided by the quality factor.
00:09:04.650 --> 00:09:12.650
This provides a pretty good way to determine 𝑄 since just by looking at the graph we can determine the width of the peak and also its frequency, which is the resonant frequency.
00:09:13.060 --> 00:09:18.800
Furthermore, if we know any three of the quantities appearing in our full formula, we can use that formula to find the fourth.
00:09:19.240 --> 00:09:24.300
Alright, now that we’ve learned about the resonant frequency and the 𝑄-factor, let’s work through some examples.
00:09:24.780 --> 00:09:29.390
A circuit consists of a resistor, a capacitor, and an inductor all of which are in series.
00:09:29.610 --> 00:09:33.960
An alternating-voltage source is connected to the circuit, and an alternating current is generated.
00:09:34.300 --> 00:09:38.650
How does the resonant frequency of the circuit change if the inductance of the inductor is increased?
00:09:39.020 --> 00:09:41.230
(a) The resonant frequency decreases.
00:09:41.350 --> 00:09:43.640
(b) The resonant frequency increases.
00:09:43.840 --> 00:09:46.280
(c) The resonant frequency does not change.
00:09:46.570 --> 00:09:50.340
The question is asking us about the resonant frequency of an alternating-current circuit.
00:09:50.800 --> 00:10:00.550
Specifically, for a series circuit with a resistor, capacitor, and inductor, the question is asking us what will happen if the inductance of the inductor is increased.
00:10:01.030 --> 00:10:02.410
Here’s a diagram of our circuit.
00:10:02.630 --> 00:10:10.650
We have the alternating-voltage source, a resistor of resistance 𝑅, an inductor of inductance 𝐿, and a capacitor of capacitance 𝐶.
00:10:11.070 --> 00:10:14.930
We’ll use the symbol 𝜔 for the angular frequency of the voltage source.
00:10:15.130 --> 00:10:22.430
Recall that resonance will occur in this circuit when the difference between inductive and capacitive reactances, that is, the total reactance, is zero.
00:10:22.700 --> 00:10:26.370
In other words, resonance is when the inductive and capacitive reactances are equal.
00:10:26.830 --> 00:10:37.870
We also have the formulas that relate angular frequency to reactance as inductive reactance is angular frequency times inductance and capacitive reactance is one divided by angular frequency times capacitance.
00:10:38.060 --> 00:10:48.060
If we equate these expressions, as will be true at the resonant frequency, we get that 𝜔 naught 𝐿 is equal to one divided by 𝜔 naught 𝐶, where 𝜔 naught is the resonant angular frequency.
00:10:48.320 --> 00:10:57.840
If we solve this equality for 𝜔 naught, we find that the resonant angular frequency is equal to one divided by the square root of the inductance of the inductor times the capacitance of the capacitor.
00:10:58.200 --> 00:11:02.620
This formula relates resonant frequency to inductance, so let’s use it to answer our question.
00:11:03.050 --> 00:11:07.570
As inductance is increased, the square root of inductance times capacitance is increased.
00:11:07.750 --> 00:11:12.920
So the denominator of our fraction is getting larger, which means the value of the overall fraction is getting smaller.
00:11:13.370 --> 00:11:16.130
But the value of this fraction is just the resonant frequency.
00:11:16.370 --> 00:11:20.820
So as the inductance of the inductor increases, the resonant frequency decreases.
00:11:21.060 --> 00:11:27.640
Interestingly, we can see from our formula that the resonant frequency would also decrease if we increase the capacitance of the capacitor.
00:11:27.900 --> 00:11:32.050
But if we change the resistance of the resistor, the resonant frequency wouldn’t change.
00:11:32.890 --> 00:11:36.560
Let’s now see another example that deals with resonance in a more quantitative way.
00:11:37.500 --> 00:11:40.600
What is the resonant frequency of the circuit shown in the diagram?
00:11:40.870 --> 00:11:51.240
The circuit consists of an alternating-voltage source connected to a series combination of a 35-Ω resistor, a 7.5-henry inductor, and a 350-microfarad capacitor.
00:11:51.490 --> 00:11:54.470
And we’re asked to find the resonant frequency of this circuit.
00:11:54.750 --> 00:12:00.050
Recall that the inductive reactance in a circuit is the angular frequency of the voltage source times the inductance.
00:12:00.180 --> 00:12:05.580
And the capacitive reactance is one divided by the angular frequency of the voltage source times the capacitance.
00:12:05.730 --> 00:12:08.190
On resonance, these two reactances are equal.
00:12:08.360 --> 00:12:16.760
If we call the resonant angular frequency 𝜔 naught, then we have that 𝜔 naught 𝐿 is equal to one over 𝜔 naught 𝐶, which we can solve for 𝜔 naught.
00:12:17.080 --> 00:12:26.920
When we solve this equation for 𝜔 naught, we find that the resonant angular frequency is equal to one divided by the square root of the inductance of the inductor times the capacitance of the capacitor.
00:12:27.310 --> 00:12:32.260
Now, this is a formula for angular frequency, but we’re looking for just regular frequency.
00:12:32.470 --> 00:12:36.840
So we need to use the relationship that angular frequency is two 𝜋 times the regular frequency.
00:12:37.390 --> 00:12:42.760
Alright, so let’s plug our definition for angular frequency into our equation for the resonant angular frequency.
00:12:43.250 --> 00:12:49.690
We have two times 𝜋 times the resonant frequency is equal to one divided by the square root of the inductance times the capacitance.
00:12:49.730 --> 00:12:54.330
To get this expression into the final form we need, we simply divide both sides by two 𝜋.
00:12:54.500 --> 00:12:58.490
On the left-hand side, two 𝜋 divided by two 𝜋 is one, and we’re just left with 𝑓 naught.
00:12:58.800 --> 00:13:02.460
On the right-hand side, the two 𝜋 just becomes part of the denominator of our fraction.
00:13:02.760 --> 00:13:04.800
This leaves us with the final formula we need.
00:13:04.960 --> 00:13:12.120
Resonant frequency is equal to one divided by two 𝜋 times the square root of the inductance, in henries, times the capacitance, in farads.
00:13:12.350 --> 00:13:14.170
So now we just need to plug in values.
00:13:14.220 --> 00:13:15.780
We have an inductance in henries.
00:13:15.810 --> 00:13:17.490
It’s 7.5 henries.
00:13:17.730 --> 00:13:21.860
However, our capacitance is given in microfarads instead of farads.
00:13:21.900 --> 00:13:25.500
To convert to farads, recall that there are one million microfarads per farad.
00:13:25.600 --> 00:13:30.370
In other words, one microfarad is equivalent to 10 to the negative sixth farads.
00:13:30.500 --> 00:13:37.710
Since we have 350 microfarads, our capacitance is equivalent to 350 times 10 to the negative sixth farads.
00:13:37.970 --> 00:13:50.790
Plugging our inductance and capacitance into our formula for resonant frequency, this gives us one divided by two 𝜋 times the square root of 350 times 10 to the negative sixth farads times 7.5 henries.
00:13:51.150 --> 00:13:54.920
It turns out that the square root of one farad times one henry is one second.
00:13:55.120 --> 00:13:57.950
So we can rewrite the denominator with units of seconds.
00:13:58.320 --> 00:14:02.140
Now, one divided by seconds is the unit hertz, which is used for frequency.
00:14:02.440 --> 00:14:04.870
So now we have an expression for the resonant frequency.
00:14:05.030 --> 00:14:09.170
That’s a number times a unit hertz, which is the right unit for frequency.
00:14:09.430 --> 00:14:12.550
So now all we have to do is evaluate this number with a calculator.
00:14:12.720 --> 00:14:17.870
When we do this evaluation, we find that the entire numerical expression is approximately equal to 3.1.
00:14:18.310 --> 00:14:21.840
So the resonant frequency of this circuit is 3.1 hertz.
00:14:22.050 --> 00:14:27.070
It’s worth noting that the 35-Ω resistor played no role in our calculation of the resonant frequency.
00:14:27.790 --> 00:14:32.330
Alright, now that we’ve seen some examples, let’s review some of the key points that we’ve learned in this lesson.
00:14:32.870 --> 00:14:40.950
In this video, we considered a circuit consisting of an alternating-voltage source driving a resistor, inductor, and capacitor, all connected in series.
00:14:41.230 --> 00:14:51.620
Because inductive and capacitive reactance both depend on the frequency of the alternating-voltage source, we saw that it was possible to find a frequency where the total reactance, inductive minus capacitive reactance, is zero.
00:14:52.180 --> 00:15:05.980
By equating the frequency-dependent formulas for inductive and capacitive reactances, we were able to find that the resonant angular frequency is equal to one divided by the square root of the inductance of the inductor times the capacitance of the capacitor.
00:15:06.290 --> 00:15:15.060
When the frequency of the voltage is equal to the resonant frequency, the net effect of the inductors and the capacitors is to provide no opposition to current in the circuit.
00:15:15.340 --> 00:15:21.160
This means that the only opposition to current is from the resistor, and so the impedance is identically the resistance.
00:15:21.350 --> 00:15:28.080
This is also the minimum possible value for the impedance because off resonance, the reactive contribution to the impedance is greater than zero.
00:15:28.590 --> 00:15:33.720
Correspondingly then, by Ohm’s law, if the impedance is minimum, then the current amplitude is maximum.
00:15:33.890 --> 00:15:38.230
Finally, we looked at how the relative amplitude of the current depends on the angular frequency.
00:15:38.610 --> 00:15:45.850
We saw that a graph with relative amplitude on the vertical axis and angular frequency on the horizontal axis shows a sharp peak at the resonant frequency.
00:15:46.300 --> 00:15:54.660
This corresponds to the same magnitude of drive voltage resulting in a much larger current at the resonant frequency than at lower and higher frequencies.
00:15:55.170 --> 00:16:05.570
To quantify the sharpness of the resonant peak, we define the 𝑄- or quality factor as the resonant angular frequency times the inductance of the inductor divided by the resistance of the resistor.
00:16:05.830 --> 00:16:13.620
Larger values of 𝑄 correspond to resonances with sharper peaks, and smaller values of 𝑄 correspond to resonances with broader peaks.
00:16:13.830 --> 00:16:23.020
Finally, we stated but didn’t prove that we can determine the value of 𝑄 from a graph like this one by measuring the width of the peak and also the peak’s location, which is the resonant frequency.
00:16:23.380 --> 00:16:29.100
So by knowing any three of the values in our formula for the quality factor, we can determine the value of the fourth quantity.